我有以下应用程序:
class MyAppState extends State<MyApp>
{
TenderApiProvider _tenderApiProvider = TenderApiProvider();
Future init() async {
await _tenderApiProvider.getToken();
}
MyAppState()
{
init();
}
@override
Widget build(BuildContext context) {
return MultiProvider(
providers: [
ChangeNotifierProvider(builder: (_) => _tenderApiProvider),
],
child: MaterialApp(
title: "My App",
routes: {
'/': (context) => HomePage(),
'/splash-screen': (context) => SplashScreen(),
'/result_table': (context) => ResultDataTable(),
}
),
);
}
}
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我需要SplashScreen在开始时首先绘制当前代码显示HomePage。
在启动画面中,我需要在加载所有数据后切换到 HomePage。这是代码:
Widget build(BuildContext context) {
TenderApiProvider apiProv = Provider.of<TenderApiProvider>(context);
return StreamBuilder(
stream: apiProv.resultController,
builder: (BuildContext context, AsyncSnapshot snapshot) {
//...
if(apiProv.apiKeyLoadingState == ApiKeyLoadingState.Done && apiProv.regionsLoadingState == RegionsLoadingState.Done)
{
Navigator.of(context).pushNamed("/"); // Should it be placed in Build??
}
});
}
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你能帮我显示在应用程序启动时绘制,SplashScreen然后从它切换到HomePage吗?
您需要将您的SplashScreen()内部包装起来,StatefulWidget以便您可以在initState(). 将fetch()逻辑包装在 a 中SchedulerBinding.instance.addPostFrameCallback()以访问BuildContextinside很重要initState()。此外,这样可以避免与在实际构建时被销毁的 RenderObject 发生冲突。
遵循一个完整的最小示例。
class App extends StatelessWidget {
@override
Widget build(BuildContext context) {
return MaterialApp(
home: Wrapper(),
);
}
}
class Wrapper extends StatefulWidget {
@override
_WrapperState createState() => _WrapperState();
}
class _WrapperState extends State<Wrapper> {
@override
void initState() {
super.initState();
SchedulerBinding.instance.addPostFrameCallback((_) {
// fetch data
Navigator.of(context).pushNamed('/');
});
}
@override
Widget build(BuildContext context) {
return Scaffold(
body: SplashScreen(),
);
}
}
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